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# Class Note for ECE 6340 with Professor Jackson at UH

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Date Created: 02/06/15

ECE 6340 Intermediate EM Waves Fall 2008 Prof David R Jackson Dept of ECE Notes 1 Maxwell s Equations 5t Vm EJ WmZET U Cmz J m Am 6t Vx g Ampere39s Law Vx z Faraday39s Law V pv Gauss39s Law V 0 magnetic Gauss39s Law Current Density Current Density A small surface is perpendicular to the direction of current flow X IN dlzgds dS ml Current Density cont A more general case of arbitrary orientation zZzzamp ii surface S component of current density crossing dS Continuity Equation From Gauss s Law Vpv gt V Vja v From Ampere s Law Vx 66 80 vvx v vgj Continuity Equation cont 0 V V Va j gt Vj V at at Hence Continuity Equation From last slide apv V Integral Form of Continuity Equation 0m VIVZdV Jar dV S closed Apply divergence theorem 3gt JVZ dV ngds uxper un vohnne net ux outofS 9 Integral Form of Continuity Equation cont A 8 CjZgdS Jar dV Assume Vis stationary a d d j dV E ijdV get V V Hence ltigt charge conservation m Integral Form of Continuity Equation cont Charge that enters the region must stay there and this results in an increase of total charge inside the region 11 Generalized Continuity Equation Assume S is moving eg expanding dt Note 6199quot i pv xeny dS S f difference in charge and surface velocities 12 Generaliized Continuity Equation Define the current density in the moving coordinate system 13 Integral Forms of Maxwell s Eqs 8 Vx fz Z a Stokes s Theorem C Vx d5 39 S cireulariron per L unlt area 01 S circulation on boundary of S 14 lntegral Forms of Maxwell s Eqs cont Integrate Ampere s law ngy dS lg am dS S S S at Apply Stokes s theorem J g d5 dS C S S at or ri dS C The current is is the current though the surface S 15 Integral Forms of Maxwell s Eqs cont Ampere s law Note In statics is is independent of the shape of S since the LHS is Similarly Faraday s law Note In statics the voltage drop around a closed path is always zero the voltage drop is unique 16 Boundary Form of Maxwell s Eqs I I We allow for a surface current density The I 39 39 narrow rectangular path C is chosen arbitrarily i lll If I l oriented in the direction ras shown l z s quotl C I At Evaluate the LHS of Ampere s law on the path shown 0 lim J lim J dr r dr Boundary Form of Maxwell s Eqs cont J RHS of Ampere s law for path AT 027827LI2 A is1 1 13 gx gsd AI 3gt isampgtltiazmi altgtlt Also lim 2d520 5 0S at 18 Boundary Form of Maxwell s Eqs cont quot IIH quotquot IIIIIII H A i h I In L ilt gt2sgtlt2gt I we Since 5 is an arbitrary unit tangent vector 12t gvxllEl gtlt 13gtltggtlt13g points into region 1 19 Boundary Form of Maxwell s Eqs cont H W L h I a i r Juli L 3 j Similarly from Faraday39s Law 20 Boundary Form of Maxwell s Eqs cont A surface Charge density is allowed 39 1 l J l lull 11139 I lit tgr39 m 5 I l A pillbox surface is Chosen 5 AS 02 82Hu2 Evaluate LHS of Gauss s law over the surface shown 0 lim Q SdSlim J dS Q dd5 6 gt0S 6 gt0 S S7 55 3amp z rM9 S 2 Boundary Form of Maxwell s Eqs cont lull l RHS of Gauss s law i I l 39 S AS lim QM 1im j ps dS 02822 5 0 5 0 AS lim QMl z psAS 5 gtO Hence Similarly from the magnetic form of Gauss39s Law 22 Summary of Maxwer Equations Point Form Integral Form Boundary Form 692 692 A A ng if Jfri ndS nfo m sf wM Vx gS l Em ampx 29 at C S at V pv c wds Q9 Ql szps S V 0 qg dS o El 2M S 23 IfS is stationary A Sat gd51jnd5 dt Faraday s Law A S w magnetic flux through S 24 Faraday s Law cont If S is moving Previous form is still valid However Iggg A d A ell W 25 Faraday s Law cont Vector identity for a moving path d A 5 EJQQQISZJE A QdS C ysx i C See appendix for derivation Faraday s Law cont 893A Startwith Qz gdS C S 8t Then use ij d52 I dS C ysx y sz S at C Hence I d W q2s C 27 Define Faraday s Law cont 28 Two Forms of Faraday s Law Faraday s law Generalized Faraday s law 29 Two Forms of Ampere s Law Generalized Ampere s law 30 Example Note tis in s 0 is in m Find the voltage drop WU around the closed path Find the electric field on the path Find the EMF drop around the closed path Find the force on a charge q that is on the path Practical note At a low frequency the magnetic field is slowly varying approximately constant in the radial direction 31 Example cont Voltage drop from Faraday s law W zra A S C at j a ds S 8t 39 8t 6 Z 8t pz amtz sin wt 0 2005 wt ptt d5 S Would the answer Change if the path were not moving No 32 Example cont Electric field from voltage drop 6 2750 2 WM sin wt 2 go wsinwt Zp wtz s1nat 2p at Note There is no 0 or 2 component from the curl of the magnetic field 33 Example cont EMF from Generalized Faraday s law y 7th 003601 34 Example cont w 712 coscot path moving 61 9 2m cos mt d1 7272 sin wt field Changing EMF 2 d dt 35 Force from EMF Example cont dw dt dlt dl L27n cos wt ath sin wt 27H 36 Example cont Note The force is purely in the azimuth direction since the electric field is and so is the velocity crossed with the magnetic field Example cont Electric field alternative method from Generalized Faraday s law S 27rp DV m 2 d V 2727 vsp d l Since vSp 1 ms lt2Epgt g2gt d W Z 38 Example cont 1 dl 9 r 0050 2m dt cosat L27rtcos wt ztzwsin wt 27H wt cos wt coswt 751nmt gsin wt Vm 39 Example Summary Voltage drop WU around the closed path W0 2 amtz sin wt V Amt Electric field on the path sin wt Vm EMF drop around the closed path EMF 2 27zt COS wt amtz Sin at V A q 27 00560 Force on a charge q that IS on the path g 2 N k 27 amt sin cot 40 Example Find the voltage on the voltmeter W m y a I x voltmeter 5 0 2005 wt V0 41 Example cont 8t 2 ama2 sin wt Wm V0 2 mm2 sinat W m y a E V0 x voltmeter 5 0 cosat 42 Example Find the voltage Wm on the voltmeter Q Wm voltmeter sliding bar perfect conductor I velocity v0 39 43 Example cont EMF4ampxr C dt wW ally 11 dW I velocnty v0 d zdquot 232 222 25 0 quotVquot Lv0 voltmeter g g g g g g 3 EMF Lv0 L 44 Example cont EMFCf 2Sgtlt szO V0O l PEC wires AEMF R1 R O Hence velocity v0 1 V0 LV0 I 39 30 quotVquot voltmeter t l h TimeHarmonic Representation Assumefrz is sinusoidal fzrAzcoswr z Re A r 61 Eefwf from Euler s identity eZCOSZjSIHZ Denote F Aej r Phasor form off TimeHarmonic Representation cont f U 142ltgtOS60t Notation ft lt gt F for scalars m lt gt F for vectors 47 TimeHarmonic Representation cont Derivative Property RCFltzgteW R6 Z ej39wt RejaF 6W HJ phasor for the derivative Hence 48 TimeHarmonic Representation cont Consider a differential equation such as Faraday s Law VX ZX a f t or magma Rejw ze or ReVgtltEja ejm 9 49 TimeHarmonic Representation cont Let C Cr jci VXEjw xyz Then Recej 0 Does this imply that c 0 Phasor choose wt 0 C 0 interpretation 66ij Im choose wtzyz2 3120 Re Hence 0 0 50 TimeHarmonic Representation cont Therefore VxEja xyyz 0 SO Vx ja Q 51 TimeHarmonic Representation cont We can also reverse the process WEwa 1 ReVgtltEja ejm 1 WRergmw Rejw2er 1 Vx rt 9 8 5t 81 52 TimeHarmonic Representation cont Hence in the sinusoidal steadystate these two equations are equivalent 53 Maxwelll s Equations in TimeHarmonic Form 54 Continuity Equation TimeHarmonic Form 8p V v Z a surface 2D I is the current in the direction I 55 Frequency Domain Curl Equations At nonzero frequency the frequency domain curl equations imply the divergence equations Vx zlij VgtltE jw O gt Vx V1jaVQ JwpvVQ Wx MV39E Hence we often consider only the curl equations in the frequency domain 56 Time Averaging of Periodic Quantities 1 T Define d W T CW r Tperiod 5 Assume a product of sinusoidal waveforms ftACOSdZ0 Q FZAeja 8ZBcosat 9 GZBem ftgtABcosatacosat ZABECOSa cos2aza 57 Time Average cont ftgtABcosa cos2ata The timeaverage of a constant is simply the constant The timeaverage of a sinusoidal wave is zero Hence iABCOS0t 3 58 Time Average cont The phasors are denoted as FAej0l G236 Consider the following FG ABeM so ReFG ABcosa 8 Hence The same formula extends to vectors as well 59 Example Stored Energy Density ReDxE ReDyEReDZE or lt ReDE Example Stored Energy Density cont Similarly Example Stored Energy Density cont 62 Example Power Flow g x Jf instantaneous Poynting vector ltygtlt x gt ReltExH Define complex Poynting vector VAm2 This formula gives the timeaverage power flow Note In the figure 52 3 drawn as planar for si mplicity IAZMaauauaa Za Proof of Moving Surface Identity Append ix 63 64 Proof of Moving Surface Identity cont Hence S dszEgiiif aAt t dA tAt ds SO if dS dsmmi Hmy ds sz SO 6 AHOAZAS For the last term Hmy dsz AS Proof of Moving Surface Identity cont Examine term Iggygds d5 435 At rmds z gm wJi Ar 66 Proof of Moving Surface Identity cont A Since x gs only has an E component we can write A A t gtltrs t gtlt2s Hence tgdS z tdrgtltv At Therefore summing all the dS contributions 1 A itirlgi l dk il lmv 67 Proof of Moving Surface Identity cont Therefore d A 8 A w dS dS w d dt 5L 2 SO at 2 lt 1 Vector identity 73 air X V 2 V X dr Hence 68 Proof of Moving Surface Identity cont Note If the surface S is nonplanar the result still holds since the magnetic flux through any two surfaces that end on C must be the same since the divergence of the magnetic field is zero from the magnetic Gauss law Hence the surface can always be chosen as planar for the purposes of calculating the magnetic flux through the surface

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